This means that there are 7 protons in the nucleus of a nitrogen atom.
<h3>Answer:</h3>
Excess Reagent = NBr₃
<h3>Solution:</h3>
The Balance Chemical Equation for the reaction of NBr₃ and NaOH is as follow,
2 NBr₃ + 3 NaOH → N₂ + 3 NaBr + 3 HBrO
Calculating the Limiting Reagent,
According to Balance equation,
2 moles NBr₃ reacts with = 3 moles of NaOH
So,
40 moles of NBr₃ will react with = X moles of NaOH
Solving for X,
X = (40 mol × 3 mol) ÷ 2 mol
X = 60 mol of NaOH
It means 40 moles of NBr₃ requires 60 moles of NaOH, while we are provided with 48 moles of NaOH which is Limited. Therefore, NaOH is the limiting reagent and will control the yield of products. And NBr₃ is in excess as some of it is left due to complete consumption of NaOH.
Answer:
Potassium sulfate
Explanation:
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Answer:
I(Br)
Explanation:
You will be combining one iodine atom with one bromine atom.
Answer:
5.56 × 10⁻⁸
Explanation:
Step 1: Given data
- Concentration of the weak acid (Ca): 0.187 M
Step 2: Calculate the concentration of H⁺
We will use the following expression.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M
Step 3: Calculate the acid dissociation constant (Ka)
We will use the following expression.
![Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%20%7D%7BCa%7D%20%3D%20%5Cfrac%7B%281.02%20%5Ctimes%2010%5E%7B-4%7D%29%5E%7B2%7D%20%7D%7B0.187%7D%20%3D%205.56%20%5Ctimes%2010%5E%7B-8%7D)
